Harmonic Function on Disk with Discontinuity on Boundary This is an old qualifying exam problem that I can't solve:
Given bounded harmonic function $h(z)$ on $\mathbb{D}$ with limits s.t.  $\lim_{r \to 1^-} h(re^{it})$ equals 1 for $0<t<\pi$ and equals 0 for $\pi<t< 2\pi$. Find $h(\frac{1}{2})$.
This seems mysterious and I don't know where to start.
 A: The answer is that $h(1/2) = 1/2$.
This follows from symmetry.  Specifically, the function $h(z)$ is uniquely determined in the interior and, because of the symmetry of the boundary conditions, must satisfy the functional equation
$$
h(\overline{z}) = 1 - h(z).
$$
(To prove this equation, observe that $h(z) + h(\overline{z})$ is harmonic and is equal to $1$ almost everywhere on the boundary.)  It follows that $h(z) = 1/2$ for values of $z$ in the interior of the disk that lie on the real axis.
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\ic}{{\rm i}}%
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 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{{\rm h}\pars{r,\theta} = \half + \fermi\pars{r,\theta}}$ where $\ds{\fermi\pars{1^{-},\theta} = \half}$ when $\ds{0 < \theta < \pi}$ and $\ds{-\,\half}$ when $\ds{\pi < \theta < 2\pi}$. Then, $\ds{\fermi\pars{r,\theta} = -\fermi\pars{r,2\pi - \theta}}$.
$\ds{\fermi\pars{r,\theta} = \sum_{n = 1}^{\infty}A_{n}r^{n}\sin\pars{n\theta}}$:
\begin{align}
\half = \sum_{n = 1}^{\infty}A_{n}\sin\pars{n\theta}\quad\imp\quad&
\half\int_{0}^{\pi}\sin\pars{m\theta}\,\dd\theta
=
\sum_{n = 1}^{\infty}A_{n}
\int_{0}^{\pi}\sin\pars{m\theta}\sin\pars{n\theta}\,\dd\theta
\\[3mm]&=
A_{m}\int_{0}^{\pi}\sin^{2}\pars{m\theta}\,\dd\theta = A_{m}\,{\pi \over 2}
\end{align}

\begin{align}
A_{n} &= {1 \over \pi}\int_{0}^{\pi}\sin\pars{n\theta}\,\dd\theta
= {1 \over \pi}{-\cos\pars{n\pi} + 1 \over n}
= {1 \over \pi}{-\pars{-1}^{n} + 1 \over n}
=
\left\lbrace%
\begin{array}{lcl}
{2 \over n\pi} & \mbox{if} & n\ \mbox{is odd}
\\[2mm]
0 & \mbox{if} & n\ \mbox{is even}
\end{array}\right.
\end{align}

$$
\fermi\pars{r,\theta}
=
{2 \over \pi}
\sum_{n = 0}^{\infty}{r^{2n + 1}\sin\pars{\bracks{2n + 1}\theta}
\over 2n + 1}
=
{2 \over \pi}\Im
\overbrace{\sum_{n = 0}^{\infty}
{z^{2n + 1} \over 2n + 1}}^{\ds{\equiv\ \varphi\pars{z}}}\,,
\qquad z \equiv r\expo{\ic\theta} \equiv x + y\ic\,,\quad x, y\ \in\ {\mathbb R}
$$

$\ds{\varphi'\pars{z} = \sum_{n = 0}^{\infty}z^{2n} = {1 \over 1 - z^{2}}
= {1 \over 2\pars{1 - z}} + {1 \over 2\pars{1 + z}}\,,\quad\varphi\pars{0} = 0}$
\begin{align}
&\Im\varphi\pars{z} = \half\Im\ln\pars{1 + z \over 1 - z}
= \half\Im\ln\pars{\pars{1 + z}\pars{1 - z^{*}} \over \verts{1 - z}^{2}}
=\half\Im\ln\pars{1 + z - z^{*} - \verts{z}^{2}}
\\[3mm]&=\half\Im\ln\pars{1 - r^{2} + 2y\ic}
=\half\,\arctan\pars{2y \over 1 - r^{2}}=
\half\,\arctan\pars{{2r \over 1 - r^{2}}\,\sin\pars{\theta}}
\end{align}

$$\color{#0000ff}{\large%
{\rm h}\pars{r,\theta}
=
\half  + {1 \over \pi}\,\arctan\pars{{2r \over 1 - r^{2}}\,\sin\pars{\theta}}}
$$
